= Solution
The <parity operator> acts by $(P\psi)(x)=\psi(-x)$. For an even potential, differentiating twice and using $V(-x)=V(x)$ proves $HP=PH$ on the parity-invariant domain of the <Hamiltonian operator>. Therefore each <energy eigenspace> is invariant under $P$. The projectors $(I+P)/2$ and $(I-P)/2$ split it into even and odd subspaces, giving a basis of simultaneous <eigenstates>. An odd state obeys $\psi(0)=-\psi(0)$, hence \b[$\psi_{\rm odd}(0)=0$].
The infinite walls impose \b[$\psi(-a)=\psi(a)=0$], with the <wavefunction> zero outside. It is continuous at the central <delta potential>. Integrate the stationary <Schrödinger equation> over $[-\varepsilon,\varepsilon]$; the energy <integral> tends to zero and the <Dirac delta function> contributes $\kappa\psi(0)$. This gives
$$
\boxed{\frac{\hbar^2}{2m}\bigl(\psi'(0+)-\psi'(0-)\bigr)=\kappa\psi(0).}
$$
For the given piecewise positive-energy form, $E=\hbar^2\lambda^2/(2m)$. Replacing $x$ by $-x$ swaps the two pieces and leaves their values unchanged, proving $P\psi=\psi$. The value at zero is $A$, and the one-sided <derivatives> are $-B\lambda$ and $B\lambda$. The jump condition gives $B=-m\kappa A/(\hbar^2\lambda)$. The wall at $a$ gives $A\cos(\lambda a)-B\sin(\lambda a)=0$, hence
$$
\boxed{\tan(\lambda a)=-\frac{\hbar^2\lambda}{m\kappa}.}
$$
For $\kappa=0$ the undivided wall equation instead gives the usual free even roots. For $\kappa>0$, set $c=m\kappa/\hbar^2>0$ and $k_n=(n+1/2)\pi/a$. In every interval $k_n<\lambda<(n+1)\pi/a$ there is exactly one root: writing $\delta=a(\lambda-k_n)\in(0,\pi/2)$ gives $\cot\delta=\lambda/c$, whose two sides are respectively strictly decreasing and increasing. These are all positive even roots; the quadratic form $\hbar^2\int|\psi'|^2/(2m)+\kappa|\psi(0)|^2$ is nonnegative, so there are no negative <eigenvalues>. Thus $\lambda_n>k_n$ and \b[$\eta_n=\hbar^2(\lambda_n^2-k_n^2)/(2m)>0$].
The printed limit of this absolute energy difference is false. The exact quantization equation gives $\delta_n=\arctan(c/\lambda_n)$, so $\delta_n\to0$ but $\lambda_n\delta_n\to c$. Consequently
$$
\lambda_n^2-k_n^2=\frac{2\lambda_n\delta_n}{a}-\frac{\delta_n^2}{a^2}\longrightarrow\frac{2c}{a},\qquad \boxed{\lim_{n\to\infty}\eta_n=\frac{\kappa}{a}.}
$$
Thus the <high-energy shift from a central delta barrier> is a finite positive constant, not zero. The corrected vanishing statement is the relative shift $\eta_n/E_n^{(0)}\to0$, or the wave-number difference $\lambda_n-k_n\to0$. Physically, highly energetic particles are weakly affected in relative terms by a fixed point barrier. The free even <wavefunction> normalized in $[-a,a]$ has density $1/a$ at the origin, so the barrier's first-order expectation is $\kappa/a$, matching the exact limit.
Odd states vanish at zero, so the delta term and the <derivative> jump both vanish. Their normalized <wavefunctions> and energies are
$$
\boxed{\psi_j(x)=a^{-1/2}\sin(j\pi x/a)\quad(|x|<a),\qquad E_j^{\rm odd}=\frac{\hbar^2\pi^2j^2}{2ma^2},\quad j=1,2,\ldots,}
$$
with zero <wavefunction> outside. They do not depend on $\kappa$. These even and odd sectors describe the <central delta barrier in a symmetric infinite well>.
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